简单，二分法，可是在oj上交的时候会有精度的限制，所以仅仅能把数扩得大一点，并且在扩的时候为防止尾数会自己主动生成一些非零数，所以还是自己吧扩到的位置设置为0，还有输出时由于%.2lf会自己有4设5入，所以能够自己算出小数点后两位的数值，在输出，保证要求的精度

#include 
     
      #include 
      
       #include 
       
        #define MAX 10010using namespace std;long long input[MAX];int m,n;bool cmp(int a,int b){    return a
        
         < n;i++){        x += input[i] / mid;    }    if(x >= m)        return true;    else        return false;}int main(){//#ifndef ONLINE_JUDGE  //  freopen("in.txt","r",stdin);//#endif    long long high,low,mid,res;    double a;    while(cin>>n>>m){        high = 0;        low = 0;        for(int i = 0;i < n;i++){            cin>>a;            input[i] = (a+1e-8)*10000;            high = high > input[i]?high : input[i];        }        sort(input,input+n,cmp);        res = 0;        low = 10000/100;        while(low <= high){            mid = (low+high)/2;            if(judge(mid)){                res = mid;                low = mid+1;            }            else                high = mid-1;        }        int th = res/10000;        int hu = (res - th * 10000)/1000;        int te = (res - th * 10000 - hu * 1000)/100;      printf("%d.%d%d\n",th,hu,te);    }}